JonathanBr
New member
- Local time
- Today, 17:19
- Joined
- Aug 31, 2017
- Messages
- 7
Hello
I am newbie and really hope someone can assist.
I setup a "main" database with related tables. But there is one table with lots of information based on a primary field (a "curve"). I did create an autonumber field to drive new instances of curves.
A form was created with a lot of information uniquely related to that autonumber. However there was one field that has caused problems. We have an image for each unique curve, so it seemed a good idea to add an "attachment" type of field to store the jpeg. As it is unique one-to-one relationship, I thought that it would be fine a single field. And initially it is.
However, we want to have the data available to a different (secondary) database, via a linked table or query. This is where problems arise. If possible, my boss wants the data in the secondary database to be read only - which could be done with a query in the second database,, linking straight to the main database.
This has caused problems. As a newbie I did not know the attachment field is a Multi-value field. A table with a multi-value field cannot be referenced in an insert into query in another database. But if the attachment field is removed from the actual main database, that query will work.
But in the data entry form in the main database, I need to allow the jpeg to be referenced and added in. I will of course add a new table to the link - give it a field number and relate it to the autonumber in the existing table. But I'd like to not have a subform, (as the existing form is a split form anyway and works well for users. Furthermore, it seems pointless in creating a subform for 1 field. Afterall it has a one-to-one relationship with records in the main table.).
So I could create a multi-table select query to feed the form. that will be fine. But is that it? When I press the add button, will the same autonumber be populated in the main table and the related table for the images. Or do I need to do something with VBA to make that work?
Or will I need a subform?
Any help be much appreciated.
regards
I am newbie and really hope someone can assist.
I setup a "main" database with related tables. But there is one table with lots of information based on a primary field (a "curve"). I did create an autonumber field to drive new instances of curves.
A form was created with a lot of information uniquely related to that autonumber. However there was one field that has caused problems. We have an image for each unique curve, so it seemed a good idea to add an "attachment" type of field to store the jpeg. As it is unique one-to-one relationship, I thought that it would be fine a single field. And initially it is.
However, we want to have the data available to a different (secondary) database, via a linked table or query. This is where problems arise. If possible, my boss wants the data in the secondary database to be read only - which could be done with a query in the second database,, linking straight to the main database.
This has caused problems. As a newbie I did not know the attachment field is a Multi-value field. A table with a multi-value field cannot be referenced in an insert into query in another database. But if the attachment field is removed from the actual main database, that query will work.
But in the data entry form in the main database, I need to allow the jpeg to be referenced and added in. I will of course add a new table to the link - give it a field number and relate it to the autonumber in the existing table. But I'd like to not have a subform, (as the existing form is a split form anyway and works well for users. Furthermore, it seems pointless in creating a subform for 1 field. Afterall it has a one-to-one relationship with records in the main table.).
So I could create a multi-table select query to feed the form. that will be fine. But is that it? When I press the add button, will the same autonumber be populated in the main table and the related table for the images. Or do I need to do something with VBA to make that work?
Or will I need a subform?
Any help be much appreciated.
regards